# Lab 4
Name: G Jaswanth
Roll No.: CS22B020
---
## Question 1
**Code**
```assembly=
.data
array: .word 1 2 3 4 5 6 7 8 9 10 11
str: .string " "
.text
la x1 array
addi x2 x0 3 #storing s=3
addi x3 x0 11 # iterator i=10
addi x1 x1 40 # moving the starting addr to array + 40 times
addi x7 x7 11
loop:
lw x4 0(x1) #loading the address into x4
add x4 x4 x2 #adding x[i] = x[i] +3
sw x4 0(x1) #storing the incremented value
addi x3 x3 -1 # i-- since started with i=10
addi x1 x1 -4 # since we are at the last addr of array we need to decrement the addr
beq x3 x0 print #if i==0 start printing
j loop
print:
addi x1 x1 4
lw x4 0(x1)
addi a0 x4 0 #printing the incremented values
li a7 1
ecall
la a0 str # printing space between integers
li a7 4
ecall
addi x3 x3 1 # increasing the iterator i++
beq x3 x7 Exit # if i==11 it gets exit
j print
Exit: # exit from the code
li a7 10
ecall
```
_You can use either one of these according to the question_
**1 a**
output:
4 5 6 7 8 9 10 11 12 13 14
---
## Question 2
**Code**
```assembly=
.data
array: .word 1 4 3 5 7 8
str1: .string " True, there exists 3 consecutive odd nums"
str2: .string " False, there is no 3 consecutive odd nums"
.text
la x1 array #loading array addr into x1
addi x2 x0 0 # int i=0
addi x3 x0 6 # array size n=6
addi x20 x3 -2
loop:
lw x4 0(x1) #loading all arr[i]
lw x5 4(x1) #loading arr[i+1]
lw x6 8(x1) #loading arr[i+2]
addi x7 x0 1 # storing 1 in x7
and x8 x7 x4 # checking condn whther arr[i] is odd
and x9 x7 x5 # checking condn whther arr[i+1] is odd
and x10 x7 x6 # checking condn whther arr[i+2] is odd
#below code is for whther there are 3 consec or not
and x8 x8 x9
and x8 x8 x10
beq x8 x7 print1 # if condn satisfy it prints true
addi x2 x2 1 # i++
addi x1 x1 4 # increasing arr[i] to arr[i+1]
beq x2 x20 print2 # if no 3 consecutive it prints false
j loop
print1:
la a0 str1 # to print true
li a7 4
ecall
li a7 10
ecall
print2:
la a0 str2 # to print false
li a7 4
ecall
```
```cpp=
class Solution {
public:
bool threeConsecutiveOdds(vector<int>& arr) {
int n=arr.size();
int flag=0;
for(int i=0;i<n-2;i++){
if(arr[i]%2==1 && arr[i+1]%2==1 && arr[i+2]%2==1){
flag=1;
break;
}
}
if(flag==1)
return true;
return false;
}
};
```
_You can use either one of these according to the question_
**2 a**
Output:
True, there exists 3 consecutive odd nums
---
## Question 3
**Code**
```assembly=
.data
array: .word 1,2,3,4,5,11,12,13,14,15
str: .string " "
newarray: .word 0x10000000
.text
addi x20 zero 10 # size of the array
la x1 array
la x5 array
la x2 newarray
la x3 newarray
addi x4 zero 0
addi x29 x29 0
loop:
beq x20 x4 print # 10!=0
lw x7 0(x1) #loading arr[i] in x7
lw x8 20(x5) #loading the arr[n+i] in x8
sw x7 0(x2) #storing the loaded words
sw x8 4(x2)
addi x1 x1 4
addi x5 x5 4
addi x2 x2 8
addi x4 x4 2
j loop
print:
lw x15 0(x3) #load the address
beq x20 x29 Exit #exiting if they are e
addi x29 x29 1 #increment the x29
addi x3 x3 4
addi a0 x15 0 #for printing changed values
li a7 1
ecall
la a0 str # for printing space
li a7 4
ecall
j print
Exit:
li a7 10 # to Exit
ecall
```
```cpp=
class Solution {
public:
vector<int> shuffle(vector<int>& nums, int n) {
vector <int> v;
for(int i=0;i<n;i++){
v.push_back(nums[i]);
v.push_back(nums[i+1]);
}
return v;
}
};
```
_You can use either one of these according to the question_
**3 a**
1 11 2 12 3 13 4 14 5 15
**Observation:**
* We gave input as
* 1 2 3 4 5 11 12 13 14 15
* and output as 1 11 2 12 3 13 4 14 5 15 by taking two pointers at i and i+n position.
---
## Question 4
**Code**
```assembly=
.data
array:.word 5,2,7,5,8,3,9,6
base:.word 0x10000000
str:.string " "
.text
addi x12 x0 8 #size of array int
lw x14 base
addi x15 x0 5 # we need to remove this element
Loop:
beq x18 x12 exit
lw x20 0(x14)
addi x14 x14 4
addi x18 x18 1 #i++
bne x20 x15 print # if it is not equal to selected elemt it goes to print
j Loop
print:
li a7 1 # for printing except the selected element
addi a0 x20 0
ecall
li a7 4 #For printing spaces
la a0 str
ecall
j Loop
exit:
li a7 10 #Exit
ecall
```
```cpp=
class Solution {
public:
void Print_Emnts(vector<int>& nums, int n) {
for(int i=0;i<n;i++){
if(nums[i]!=n){
cout<<nums[i]<<" ";
}
}
}
};
```
_You can use either one of these according to the question_
**4 a**
Output:
2 7 8 3 9 6
**Observation:**
This code is to print the array elemnts except the selected value..
---
## Question 5
**Code** (if required)
```assembly=
.data
newline: .string "\n"
.text
main:
# Initialize the linked list
la x5, elem1 # Load the address of the first element into x5
la x6, elem2 # Load the address of the second element into x6
la x7, elem3 # Load the address of the third element into x7
la x8, elem4 # Load the address of the fourth element into x8
# Link the elements together
sw x6, 4(x5) # Store the address of elem2 in the next pointer of elem1
sw x7, 4(x6) # Store the address of elem3 in the next pointer of elem2
sw x8, 4(x7) # Store the address of elem4 in the next pointer of elem3
# Print all elements of the linked list
la a0, elem1 # Load the address of the head of the list into a0
jal ra, print_list
# Insert a new element into the linked list
la x9, new_elem # Load the address of the new element into x9
lw x10, 4(x5) # Load the next pointer of elem1 into x10
sw x9, 4(x5) # Store the address of the new element in the next pointer of elem1
sw x10, 4(x9) # Store the address of elem2 in the next pointer of the new element
# Print all elements of the updated linked list
la a0, elem1 # Load the address of the head of the list into a0
jal ra, print_list
# Exit program
li a7, 10 # Exit syscall
ecall
print_list:
# Load the address of the head of the list into a1
mv a1, a0
loop:
# Load the data of the current node into a1
lw a1, 0(a1)
# Print the value of the current node
mv a0, a1 # Set a0 to the data value for printing
li a7, 1 # Print integer syscall
ecall
# Print a newline character
la a0, newline # Load the newline character into a0
li a7, 4 # Print string syscall
ecall
# Get the next pointer
lw a0, 4(a0)
# Check if we reached the end of the list
bnez a0, loop # If the next pointer is not zero, continue to the next node
ret
.data
elem1: .word 1, 0 # First element of the linked list
elem2: .word 2, 0 # Second element of the linked list
elem3: .word 3, 0 # Third element of the linked list
elem4: .word 4, 0 # Fourth element of the linked list
new_elem: .word 99, 0 # New element to be inserted into the linked list
```
**Observation:**
* We can Link the two node by storing the element address in the next i+4th address where i is the addr of elemnt etc
---
## Question 6
**Code** (if required)
```assembly=
.data
.align 3
FREE_LIST: .word 0 # Pointer to the head of the free list
.text
malloc:
# Function prologue
addi sp, sp, -4 # Adjust stack pointer
sw x1, 0(sp) # Save return address (using x1 instead of ra)
# Load argument (size to allocate)
mv x11, x10 # Move size to x11 (x10 will be used as return value)
# Allocate memory using sbrk
li a7, 214 # sbrk syscall number
ecall # Call sbrk
# Move return value (new memory address) to x10
mv x10, a0 # x10 already holds the return value
# Function epilogue
lw x1, 0(sp) # Restore return address
addi sp, sp, 4 # Restore stack pointer
ret # Return from function
free:
# Function prologue
addi sp, sp, -4 # Adjust stack pointer
sw x1, 0(sp) # Save return address (using x1 instead of ra)
# Load arguments
mv a0, a0 # Assuming pointer to be freed is passed in a0
# For simplicity, we do nothing with the freed memory
# Function epilogue
lw x1, 0(sp) # Restore return address
addi sp, sp, 4 # Restore stack pointer
ret # Return from function
```
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